International Mathematical Olympiad

Let $\mathbb{R}$ be the set of real numbers. We are tasked with finding all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for any real numbers $x$ and $y$, the functional equation:

$$f(f(x)f(y)) + f(x+y) = f(xy)$$

is satisfied.

### Step 1: Checking Simple Functions

#### Case 1: Constant Function

Let's first consider the constant function $f(x) = 0$. Substituting into the equation, we get:

$$f(f(x)f(y)) + f(x+y) = f(0) + f(x+y) = 0 = f(xy)$$

Since $f(xy) = 0$ holds for all $x, y$, it satisfies the functional equation. Thus, $f(x) = 0$ is a solution.

### Step 2: Exploring Other Possibilities

To identify other forms of functions that satisfy the equation, let's impose a different assumption.

#### Case 2: Linear Solutions

Suppose $f(x) = 1 - x$. Substituting into the functional equation, we find:

$$f(f(x)f(y)) = f((1-x)(1-y)) = f(1 - x - y + xy) = 1 - (1 - x - y + xy) = x + y - xy$$

Substituting into the original equation:

$$f(f(x)f(y)) + f(x+y) = (x + y - xy) + (1 - (x+y)) = 1 - xy = f(xy)$$

Since this satisfies the functional equation for all $x, y$, $f(x) = 1 - x$ is indeed a solution.

#### Case 3: Alternate Linear Solutions

Consider $f(x) = x - 1$. Substituting into the functional equation:

$$f(f(x)f(y)) = f((x-1)(y-1)) = f(xy - x - y + 1) = xy - x - y + 1 - 1 = xy - x - y$$

The equation becomes:

$$f(f(x)f(y)) + f(x+y) = (xy - x - y) + (x + y - 1) = xy - 1 = f(xy)$$

Therefore, $f(x) = x - 1$ also satisfies the functional equation.

### Conclusion

The solutions to the functional equation are:

$$f(x) = 0, \quad f(x) = 1 - x, \quad f(x) = x - 1$$

Thus, the complete set of solutions is:
$$\boxed{f(x) = 0, \ f(x) = 1-x, \ f(x) = x-1}$$

These three functions are the only ones that satisfy the given functional equation for all $x, y \in \mathbb{R}$.